Is My Solution to This Log Question Correct?

In summary, Samy found that if x is 1 or 9, then the equation x2 - 10x + 1 = 0 has a solution of x = 1.
  • #1
elctronoob
13
3
Moved from a technical forum, so homework template missing.
Hi folks,

I'm revisiting logs for the first time in a long time through distance education and I was wondering if someone could have a look over a question I've answered and let me know if I've done it correctly or if I'm way off please

Find x if Log3(10x – 1) – 2 = 2log3x

I instantly divide by common log of log3 and work out from there

(10x-1) - 2 / 2 = x
(5x- 1/2) -1 = x
5x - 1/2 = x + 1
5x - x = 1 + 1/2
4x = 3/2
x = 3/8

Any help would be greatly appreciated
 
Physics news on Phys.org
  • #2
By 2Log3, does it mean that it's a log with basis of 2 or 2*Log3?
 
  • #3
its written as 2log3x, so I take it to mean 2 * log3X
 
  • #4
If it means
Find ##x## if ##\log_3(10x-1) -2=2*\log_3x##,
I don't understand what you did.

You could take all the ##log_3## terms together, making use of:
##\log_3(a*b)=\log_3a+\log_3b##
##\log_3(a/b)=\log_3a-\log_3b##
##2*\log_3a=\log_3a^2##
##3^{log_3a}=a##

and then solve for x.
 
  • #5
Since 2log3x = 2 (Log3 + Log x)
which means there will 2 Log 3, so there would be still one Log 3 remaining.

Or you can just actually change the 2Log3x to Log(3x)^2, and then manipulate the left side of equation to be Log, and equals them.
 
  • #6
Thanks everyone for all the help...its been far too long since I've looked at any of this sort of stuff...
I now have it at
(log310x - log31) - 2 = log3x2
Then divide out the log3
10x - 1 - 3 = x2
0 = x2 -10x + 3

I've obviously done something wrong again as this gives pretty silly results :(
 
  • #7
You've got to make (log310x - log31) - 2 into one Log3. You don't really divide out the log, but comparing the values inside the logs. They have to be equals, since the logs would result the same.
 
  • #8
elctronoob said:
Thanks everyone for all the help...its been far too long since I've looked at any of this sort of stuff...
I now have it at
(log310x - log31) - 2 = log3x2
How did you get the expression to the left?
Note that ##\log_3(10x-1)\neq\log_3(10x)-\log_3(1)##

elctronoob said:
Then divide out the log3
Not sure what you mean by "divide out the log3"
 
  • #9
You start with finding what the value of 2 is in ##log_3##

##log_3(3)=1## and ## log_3(9)=log_3(3^2)=2##

So the RHS becomes ##log_3(10x - 1) - log_3(9) = log_3( (10x - 1) * 1/9) ##

and go from there.
 
  • #10
Samy_A said:
How did you get the expression to the left?
Note that ##\log_3(10x-1)\neq\log_3(10x)-\log_3(1)##

Not sure what you mean by "divide out the log3"

i'm just going round in circles here and getting nowhere I'm afraid
##\log_3(10x-1)=((log_3(10) + log_3(x)) / log_3(1))##

I understood when i had a common log base, i could divide all by that log base
hence ((10 + x) / 1) - 2 = x2
0 = x2 - x - 8

which doesn't give reasonable answers so i must have messed up somewhere again
 
  • #11
You're not dividing by log, you have to get that understanding straight first to solve this problem:

##log3(10x−1) =/= ((log3(10)+log3(x))/log3(1))##
 
  • #13
  • Like
Likes Samy_A and jedishrfu
  • #14
elctronoob said:
i'm just going round in circles here and getting nowhere I'm afraid
##\log_3(10x-1)=((log_3(10) + log_3(x)) / log_3(1))##

I understood when i had a common log base, i could divide all by that log base
hence ((10 + x) / 1) - 2 = x2
0 = x2 - x - 8

which doesn't give reasonable answers so i must have messed up somewhere again

Right folks, I've gone away n come back with an improved effort i think...

log3(10x-1) - 2 = log3(x2)

=log3((10x-1)/x2) = 2

= (10x-1)/x2 = 32 = 9

9x2 -10x + 1 = 0

=>x = 1, x = 1/9

Does this look correct to you guys or am I missing something again?
 
  • Like
Likes Samy_A
  • #15
elctronoob said:
Right folks, I've gone away n come back with an improved effort i think...

log3(10x-1) - 2 = log3(x2)

=log3((10x-1)/x2) = 2

= (10x-1)/x2 = 32 = 9

9x2 -10x + 1 = 0

=>x = 1, x = 1/9

Does this look correct to you guys or am I missing something again?
Looks correct.
 
  • #16
Samy_A said:
Looks correct.

Thanks very much for your help Samy and the rest of you fine people, much appreciated
 

FAQ: Is My Solution to This Log Question Correct?

What is "Solving for x log question"?

"Solving for x log question" refers to a type of mathematical problem where the variable "x" is contained within a logarithmic function. The goal is to find the value of "x" that satisfies the equation.

What is a logarithmic function?

A logarithmic function is a mathematical function that represents the inverse of an exponential function. It is commonly written as logb(x), where "b" is the base and "x" is the input value. The logarithm of a number is the exponent to which the base must be raised to produce that number. For example, log2(8) = 3, since 23 = 8.

What are the steps for solving a "Solving for x log question"?

The general steps for solving a "Solving for x log question" are as follows:

  1. Isolate the logarithmic function on one side of the equation.
  2. Convert the logarithmic function to its exponential form.
  3. Solve for "x" using basic algebraic operations.
  4. Check your solution by plugging it back into the original equation.

What are some common properties of logarithmic functions?

Some common properties of logarithmic functions include:

  • The logarithm of a product is equal to the sum of the logarithms of its factors: logb(xy) = logb(x) + logb(y)
  • The logarithm of a quotient is equal to the difference of the logarithms of its numerator and denominator: logb(x/y) = logb(x) - logb(y)
  • The logarithm of a power is equal to the exponent multiplied by the logarithm of the base: logb(xn) = n*logb(x)

What are some common mistakes to avoid when solving "Solving for x log question"?

Some common mistakes to avoid when solving "Solving for x log question" include:

  • Forgetting to check for extraneous solutions, which can occur when taking the logarithm of a negative number.
  • Misapplying logarithmic properties, such as adding instead of subtracting when dealing with a quotient.
  • Not simplifying the equation before solving for "x".
  • Using the wrong base when converting to exponential form.

Back
Top